You have found the following ages (in years) of all 5 tigers at your local zoo: $ 4,\enspace 23,\enspace 5,\enspace 26,\enspace 10$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{4 + 23 + 5 + 26 + 10}{{5}} = {13.6\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-9.6$ years $92.16$ years $^2$ $23$ years $9.4$ years $88.36$ years $^2$ $5$ years $-8.6$ years $73.96$ years $^2$ $26$ years $12.4$ years $153.76$ years $^2$ $10$ years $-3.6$ years $12.96$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{92.16} + {88.36} + {73.96} + {153.76} + {12.96}} {{5}} $ $ {\sigma^2} = \dfrac{{421.2}}{{5}} = {84.24\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{84.24\text{ years}^2}} = {9.2\text{ years}} $ The average tiger at the zoo is 13.6 years old. There is a standard deviation of 9.2 years.